Chapter 10.4, Problem 75E

$\left(a\right)$

To determine

To calculate: The value of C such that the circle $x^2+y^2=4$ and the parabola $y=x^2+C$ does not intersect each other.

Answer to Problem 75E

The value of C is $C>2\text{ or }C<-\frac{17}{4}$ . The graph of the circle is provided below,

  

Explanation of Solution

Given information:

The equation of circle $x^2+y^2=4$ and the parabola $y=x^2+C$ .

Calculation:

Consider the equation of circle $x^2+y^2=4$ and the parabola $y=x^2+C$ . The graph of the circle is provided below,

  

The parabola $y=x^2+C$ is the vertical translation of the parabola $y=x^2$ . So it will not intersect the circle when the vertex of the parabola is above the circle.

It is clear from the equation of circle $x^2+y^2=4$ that its center is origin and radius is 2 units.

Therefore, the parabola and circle won’t intersect when $C>2$ .

Now, substitute $y=x^2+C$ in the equation $x^2+y^2=4$ and simplify,

  $\begin{array}{c}{x}^2+{\left(x^2+C\right)}^2=4\\ x^2+x^4+2C{x}^2+C^2=4\\ x^4+\left(2C+1\right)x^2+\left(C^2-4\right)=0\end{array}$

Now, there is no point of intersection when discriminant is less than 0 so,

  $\begin{array}{c}{\left(2C+1\right)}^2-4\left(1\right)\left(C^2-4\right)<0\\ 4C^2+4C+1-4C^2+16<0\\ 4C+17<0\\ C<-\frac{17}{4}\end{array}$

Thus, there are 0 point of intersection when $C>2\text{ or }C<-\frac{17}{4}$ .

$\left(b\right)$

To determine

To calculate: The value of C such that the circle $x^2+y^2=4$ and the parabola $y=x^2+C$ intersect each other at 1 point.

Answer to Problem 75E

The value of C is $C=2$ . The graph of the circle is provided below,

  

Explanation of Solution

Given information:

The equation of circle $x^2+y^2=4$ and the parabola $y=x^2+C$ .

Calculation:

Consider the equation of circle $x^2+y^2=4$ and the parabola $y=x^2+C$ . The graph of the circle is provided below,

  

The parabola $y=x^2+C$ is the vertical translation of the parabola $y=x^2$ . So it will not intersect the circle when the vertex of the parabola is above the circle.

It is clear from the equation of circle $x^2+y^2=4$ that its center is origin and radius is 2 units.

Therefore, the parabola and circle won’t intersect when $C>2$ .

To intersect each other at one point C must be equal to 2.

  

Thus, there is 1 point of intersection when $C=2$ .

$\left(c\right)$

To determine

To calculate: The value of C such that the circle $x^2+y^2=4$ and the parabola $y=x^2+C$ intersect each other at 2 points.

Answer to Problem 75E

The value of C is $-2<C<2\text{ or }C=-\frac{17}{4}$ . The graph of the circle is provided below,

  

Explanation of Solution

Given information:

The equation of circle $x^2+y^2=4$ and the parabola $y=x^2+C$ .

Calculation:

Consider the equation of circle $x^2+y^2=4$ and the parabola $y=x^2+C$ . The graph of the circle is provided below,

  

The parabola $y=x^2+C$ is the vertical translation of the parabola $y=x^2$ . So it will not intersect the circle when the vertex of the parabola is above the circle.

It is clear from the equation of circle $x^2+y^2=4$ that its center is origin and radius is 2 units.

Therefore, the parabola and circle won’t intersect when $C>2$ .

To intersect each other at two points C must be just below to 2 and just above $-2$ that is between the endpoints of diameter on y -axis.

Now, substitute $y=x^2+C$ in the equation $x^2+y^2=4$ and simplify,

  $\begin{array}{c}{x}^2+{\left(x^2+C\right)}^2=4\\ x^2+x^4+2C{x}^2+C^2=4\\ x^4+\left(2C+1\right)x^2+\left(C^2-4\right)=0\end{array}$

Now, there are two points of intersection when discriminant is 0 so,

  $\begin{array}{c}{\left(2C+1\right)}^2-4\left(1\right)\left(C^2-4\right)=0\\ 4C^2+4C+1-4C^2+16=0\\ 4C+17=0\\ C=-\frac{17}{4}\end{array}$

  

Thus, there are two points of intersection when $-2<C<2\text{ or }C=-\frac{17}{4}$ .

$\left(d\right)$

To determine

To calculate: The value of C such that the circle $x^2+y^2=4$ and the parabola $y=x^2+C$ intersect each other at 3 points.

Answer to Problem 75E

The value of C is $C=-2$ . The graph of the circle is provided below,

  

Explanation of Solution

Given information:

The equation of circle $x^2+y^2=4$ and the parabola $y=x^2+C$ .

Calculation:

Consider the equation of circle $x^2+y^2=4$ and the parabola $y=x^2+C$ . The graph of the circle is provided below,

  

The parabola $y=x^2+C$ is the vertical translation of the parabola $y=x^2$ . So it will not intersect the circle when the vertex of the parabola is above the circle.

It is clear from the equation of circle $x^2+y^2=4$ that its center is origin and radius is 2 units.

Therefore, the parabola and circle won’t intersect when $C>2$ .

To intersect each other at two points C must be just below to 2 and just above $-2$ that is between the endpoints of diameter on y -axis.

And it will intersect at three points when vertex of parabola touch the lower end of circle on y -axis that is when $C=-2$

  

Thus, there are three points of intersection when $C=-2$ .

  $\left(e\right)$

To determine

To calculate: The value of C such that the circle $x^2+y^2=4$ and the parabola $y=x^2+C$ intersect each other at 4 points.

Answer to Problem 75E

The value of C is $-\frac{17}{4}<C<-2$ . The graph of the circle is provided below,

  

Explanation of Solution

Given information:

The equation of circle $x^2+y^2=4$ and the parabola $y=x^2+C$ .

Calculation:

Consider the equation of circle $x^2+y^2=4$ and the parabola $y=x^2+C$ . The graph of the circle is provided below,

  

The parabola $y=x^2+C$ is the vertical translation of the parabola $y=x^2$ . So it will not intersect the circle when the vertex of the parabola is above the circle.

It is clear from the equation of circle $x^2+y^2=4$ that its center is origin and radius is 2 units.

Therefore, the parabola and circle won’t intersect when $C>2$ .

To intersect each other at two points C must be just below to 2 and just above $-2$ that is between the endpoints of diameter on y -axis.

Now, substitute $y=x^2+C$ in the equation $x^2+y^2=4$ and simplify,

  $\begin{array}{c}{x}^2+{\left(x^2+C\right)}^2=4\\ x^2+x^4+2C{x}^2+C^2=4\\ x^4+\left(2C+1\right)x^2+\left(C^2-4\right)=0\end{array}$

Now, there are two points of intersection when discriminant is 0 so,

  $\begin{array}{c}{\left(2C+1\right)}^2-4\left(1\right)\left(C^2-4\right)=0\\ 4C^2+4C+1-4C^2+16=0\\ 4C+17=0\\ C=-\frac{17}{4}\end{array}$

Now, to intersect at four points C must be below $C=-2$ and just above $C=-\frac{17}{4}$

  

Thus, there are four points of intersection when $-\frac{17}{4}<C<-2$ .